If C, equals, r, plus, 7C=r+7 and D, equals, minus, 2, r, plus, 3, commaD=−2r+3, find an expression that equals 2, C, minus, D2C−D in standard form.

Question

Solve the problem just like the example below
⬇️
If C, equals, 3, x, squared, plus, 4, x
C=3x
2
+4x and
B, equals, 2, x, squared, minus, 7, comma
B=2x
2
−7, find an expression that equals
2, C, plus, 2, B
2C+2B in standard form.
C, equals, 3, x, squared, plus, 4, x
C=
3x
2
+4x
B, equals, 2, x, squared, minus, 7
B=
2x
2
−7
2, C, plus, 2, B
2C+2B
2, left parenthesis, 3, x, squared, plus, 4, x, right parenthesis, plus, 2, left parenthesis, 2, x, squared, minus, 7, right parenthesis
2(3x
2
+4x)+2(2x
2
−7)
6, x, squared, plus, 8, x, plus, 4, x, squared, minus, 14
6x
2
+8x+4x
2
−14
10, x, squared, plus, 8, x, minus, 14
10x
2
+8x−14

Answer 1

Let's start with the given expressions for \( C \) and \( D \):

1. \( C = r + 7 \)
2. \( D = -2r + 3 \)

We need to find the expression that equals \( 2C - D \).

Step 1: Substitute values for \( C \) and \( D \)

\[ 2C - D = 2(r + 7) - (-2r + 3) \]

Step 2: Distribute

Distributing in the expression, we get:

\[ 2(r + 7) = 2r + 14 \]

And the second part becomes:

\[

• (-2r + 3) = 2r - 3 \]

Step 3: Combine the expressions

Now substituting back into the equation:

\[ 2C - D = (2r + 14) + (2r - 3) \]

Step 4: Simplify

Combine like terms:

\[ = 2r + 14 + 2r - 3 \]

\[ = (2r + 2r) + (14 - 3) \]

\[ = 4r + 11 \]

Thus, the expression \( 2C - D \) in standard form is:

\[ \boxed{4r + 11} \]